Let $\psi $ be the digamma function given by

$$\psi (z)=\left.\frac {d}{dt}\log\Gamma (t)\right|_{t=z}. $$

I wonder does anyone know of any lower and/or upper bounds on the real and imaginary parts of $\psi $? A Google search brings up some bounds for $ z=x $ real only, but I'm interested in the digamma function as a function of a complex variable. The region of interest is $0<\text{Re}(z)<1/2$.


Empirical evidence suggests $$\text{Re}(\psi(z))\approx\log(\text{Im}(z)),$$ and $$\text{Im}(\psi(z))\approx\frac{\pi}{2},$$ for $\text{Im}(\psi(z))>c$, where $c$ is "not very large". Using knowledge of $\log(z)$ for complex $z$ this leads me to believe that $$\left|\psi(z)\right|\approx \log\left|z\right|+i\text{Arg}(z)=\log z,$$ where we take the principal argument.

Indeed, after doing a bit of digging I have found that (Karatsuba, A. A., Voronin, S. M., 1991, p.344) $$\psi(z) - \log z \ll \frac{1}{\left|z\right|},$$ for $\left|\text{arg}(z)\right|<\pi$.


We have the sum form

$$ \Psi(z) = -\gamma + \sum_{k=1}^\infty \left(\dfrac{1}{k} - \dfrac{1}{k+z-1}\right) = -\gamma + \sum_{k=1}^\infty \dfrac{z-1}{k(k+z-1)}$$

If $z = x + i y$, then e.g. we have $$\text{Im} \left(\dfrac{1}{k} - \dfrac{1}{k+z-1}\right) = \dfrac{y}{(k+x-1)^2 + y^2}$$ Thus $ \text{Im}(\Psi(z))$ and $y$ have the same sign, and $$ \left|\text{Im}(\Psi(z))\right| < \sum_{j=-\infty}^\infty \dfrac{y}{(j+x)^2 + y^2} = {\frac {\pi \,\cosh \left( \pi \,y \right) \sinh \left( \pi \,y \right) }{ \left( \cosh \left( \pi \,y \right) \right) ^{2}- \left( \cos \left( \pi \,x \right) \right) ^{2}}} \le \frac {\pi \,\cosh \left( \pi \,y \right) \sinh \left( \pi \,y \right) }{ \left( \cosh \left( \pi \,y \right) \right) ^{2}- 1} $$ Lots of other bounds can be obtained. Is there any particular region of the complex plane, or asymptotic direction, you're interested in?

EDIT: For $0 \le x \le 1/2$ and $y > 0$, $\dfrac{y}{(k+x-1)^2 + y^2}$ is decreasing in $k$ for $k \ge 1$, so $$ \text{Im} \Psi(z) \ge \int_1^\infty \dfrac{y\; dt}{(t+x-1)^2 + y^2} = \dfrac{\pi}{2} - \arctan(x/y) $$ while $$ \text{Im} \Psi(z) \le \dfrac{y}{x^2+y^2} + \int_1^\infty \dfrac{y\; dt}{(t+x-1)^2 + y^2} = \dfrac{\pi}{2} - \arctan(x/y) + \dfrac{y}{x^2+y^2}$$ Similarly, although somewhat less cleanly because monotonicity is not as clear, $$ \text{Re} \Psi(z) = \sum_{k=1}^\infty \left( \dfrac{1}{k} - \dfrac{k+x-1}{(k+x-1)^2 + y^2} \right) \approx \int_1^\infty \left( \dfrac{1}{t} - \dfrac{t+x-1}{(t+x-1)^2 + y^2} \right)\; dt = \dfrac{\ln(x^2 + y^2)}{2} $$

  • $\begingroup$ Probably Euler constant $\gamma$ should be taken into consideration for $Re \Psi (z)$ $\endgroup$ – stukselbax Oct 19 '14 at 16:18

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